Leetcode: Path Sum III

You are given a binary tree in which each node contains an integer value.

Find the number of paths that sum to a given value.

The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).

The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.

Example:

root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8

      10
     /  
    5   -3
   /     
  3   2   11
 /    
3  -2   1

Return 3. The paths that sum to 8 are:

1.  5 -> 3
2.  5 -> 2 -> 1
3. -3 -> 11

Add the prefix sum to the hashMap, and check along path if hashMap.contains(pathSum+cur.val-target);

My Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int pathSum(TreeNode root, int sum) {
        if (root == null) return 0;
        ArrayList<Integer> res = new ArrayList<Integer>();
        res.add(0);
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        map.put(0, 1);
        helper(root, sum, 0, res, map);
        return res.get(0);
    }
    
    public void helper(TreeNode cur, int target, int pathSum, ArrayList<Integer> res, HashMap<Integer, Integer> map) {
        if (map.containsKey(pathSum+cur.val-target)) {
            res.set(0, res.get(0) + map.get(pathSum+cur.val-target));
        }
        if (!map.containsKey(pathSum+cur.val)) {
            map.put(pathSum+cur.val, 1);
        }
        else map.put(pathSum+cur.val, map.get(pathSum+cur.val)+1);
        if (cur.left != null) helper(cur.left, target, pathSum+cur.val, res, map);
        if (cur.right != null) helper(cur.right, target, pathSum+cur.val, res, map);
        map.put(pathSum+cur.val, map.get(pathSum+cur.val)-1);
    }
}

一个更简洁的solution: using HashMap to store ( key : the prefix sum, value : how many ways get to this prefix sum) , and whenever reach a node, we check if prefix sum - target exists in hashmap or not, if it does, we added up the ways of prefix sum - target into res.

    public int pathSum(TreeNode root, int sum) {
        Map<Integer, Integer> map = new HashMap<>();
        map.put(0, 1);  //Default sum = 0 has one count
        return backtrack(root, 0, sum, map); 
    }
    //BackTrack one pass
    public int backtrack(TreeNode root, int sum, int target, Map<Integer, Integer> map){
        if(root == null)
            return 0;
        sum += root.val;
        int res = map.getOrDefault(sum - target, 0);    //See if there is a subarray sum equals to target
        map.put(sum, map.getOrDefault(sum, 0)+1);
        //Extend to left and right child
        res += backtrack(root.left, sum, target, map) + backtrack(root.right, sum, target, map);
        map.put(sum, map.get(sum)-1);   //Remove the current node so it wont affect other path
        return res;
    }